Atwoods Analysis

[pequaide]

Any two masses can be placed in a balanced wheel and the wheel can be keep balanced. I said that if you do place them in a balanced wheel then the 200 kilograms moving .099 m/sec will accelerate the 2 kilograms moving .966 m/sec. This is because the 2 is not moving 100 times faster than the 200 kg, and in a balanced wheel that would have to be the case.

If an object is allowed to rotate freely it will rotate about its center of mass, this is also the center of momentum (linear Newtonian momentum).

Let the Atwood's accelerate both masses to .099 m/sec for the 200 kg and .966 m/sec for the 2 kg. And after these moving masses are placed in the same balanced circle or wheel then the 200 kg will drop down to 0.05433 m/sec and the 2 kg will be moving 5.433 m/sec. This is 100 times faster and the center of mass is also the center of momentum. Newtonian momentum is conserved.

The energy of the 2 kilograms is ½ * 2 kg * 5.433 m/sec * 5.433m/sec = 29.51 joules. The input energy was .1 kg dropped 1 meter twice; .981 N * 2 m = 1.96 joules

Spring don't work; they load and unload energy. There are lots of ways to do it wrong.

I am a proponent of F = ma; and all the experiments I have done prove that Newton is correct.

[Wubbly]

pequaide wrote:

Any two masses can be placed in a balanced wheel and the wheel can be keep balanced. I said that if you do place them in a balanced wheel, then the 200 kilograms moving .099 m/sec will accelerate the 2 kilograms moving .966 m/sec. This is because the 2 is not moving 100 times faster than the 200 kg, and in a balanced wheel that would have to be the case.

I am still not following you. What is your definition of a balanced wheel? Is the 200 kg @ 2 cm and the 2 kg @ 200 cm on the same wheel considered your balanced wheel? Are the masses rigidly attached to the balanced wheel?

Let the Atwood's accelerate both masses to (0.099 m/sec for the 200 kg) and (0.966 m/sec for the 2 kg). And after these moving masses are placed in the same balanced circle or wheel then the 200 kg will drop down to 0.05433 m/sec and the 2 kg will be moving 5.433 m/sec. This is 100 times faster and the center of mass is also the center of momentum. Newtonian momentum is conserved.

You seem to be mixing two different experiments. In one Atwoods experiment, 200 kg of mass is accelerated to 0.099 m/s, and in a totally separate Atwoods experiment, 2 kg of mass is accelerated to .966 m/s. Then suddenly, somehow, magically, both masses are on the same balanced wheel in some other balanced wheel experiment where one of the masses slows down and the other speeds up and energy is created? How did the 200 kg and the 2 kg magically get on the same balanced wheel in another experiment? What is the radius of the 200 kg mass on the new balanced wheel, and what is the radius of the 2 kg mass on the new balanced wheel and are they rigidly attached to the wheel? Are your velocity numbers calculated based on some conservation of pequaidian linear momentum by adding the momentum of the two atwoods experiments then evenly dividing that momentum between the two masses?

Spring don't work; they load and unload energy. There are lots of ways to do it wrong.

Yes, There are many ways to do it wrong. Springs load and unload energy. If you have an experiment that supposedly creates energy, you could determine it with a spring that stretches farther than some input baseline amount to prove energy was created. If the spring does not stretch farther than some input baseline amount, there is no extra energy in your system.

I am a proponent of F = ma; and all the experiments I have done prove that Newton is correct.

If you prove F=ma, you have proved nothing we don't already know. F=ma is not outside the known laws of physics.
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[pequaide]

I think you got it; but I don't think it would be magical to inset one Atwood's at the appropriate radius onto another rotating Atwood's. Or two pendulums swinging onto the same lever arm. But is it an engineering problem or a theoretical problem? I think it is an engineering challenge and that is some of the fun of experiments.

[pequaide]

There is a modern rotation trebuchet called ‘Inertia II’ that can throw a pumpkin 711 meters. If it is throwing at 37° then it is throwing about 260 meters up. Let’s let the missile mass be 5 kilogram and let’s make a vertical chain of 260 five kilogram masses. We could let the entire system drop 1 meter and we would have 1.3 metric tons moving 4.429 m/sec.  We could restore the system to its original configuration by throwing one 5 kilogram mass up 260 meters. A rise of 260 meters requires a velocity of 72 m/sec.

But instead of letting the 1.3 tons drop one meter lets attach it to an Atwood’s with 1.35 tons on each side. The acceleration would be 1300/4000 * 9.81 m/sec/sec = 3.188 m/sec/sec and at the end of a one meter drop the entire system would be moving 2.525 m/sec. This gives us 10,100 units of momentum. It takes about 72 m/sec * 5 kilograms = 360 units of momentum to reload the system. Lets connect the four tons to an eight cm shaft that is mounted as the hub of a 160 cm wheel. This 80 cm to 4 cm radius difference gives a 20 to 1 leverage at the circumference so the 4000 kg is now effectively only 200 kg, but this effectively 200 is now moving 50.5 meters per second. If we add 5 kilograms we are down to 205 kg moving 49.27 m/sec

If the circumference of this 160 cm wheel is wrapped with a tether and a 5 kilogram mass is thrown from the circumference, I predict that the 5 kilograms will absorb all the motion of the 4000 kilogram mass. Absorbing all of the motion does seem extreme but I have done 40 to 1 before so just going from experiment, that is what I would have to predict.

So when the missile is released, and the 4000 kg is stopped, will the 5 kilogram only increase by less than 23 m/sec to less than 72 m/sec or will it fly off like a bullet. Is the 9854 units of momentum remaining in the 4000 kilograms only going to surrender 115 units (23 m/sec * 5 kg = 115 units) before it stops? Can linear momentum disappear like this? It never has before.

Someone stated that they needed a wall to hide behind because of these violent motion changes. 49.27 m/sec to 72 m/sec don't seem like a violent motion change to me. Yet all the throws I see are violent changes in velocity.

And what of F = ma; the same force that brought 4000 kilograms to a stop is the same force that sent the 5 kilograms on its flight. In short hand we call it the Law of Conservation of Momentum; Newtonian momentum. 4000 kg *2.525 m/sec = 5 kg * X m/sec. And a wimpy 72 does not satisfy the equation.

[Wubbly]

Here is a breakdown of your "balanced wheel" with a 200 kg mass at 2 cm, and a 2 kg mass at 200 cm.

In the vertical plane, the torques (force x radius) balance each other out.

When it is rotating, the centripetal forces balance out.

But the moment of inertia of the 2 kg mass is 100 times larger than that of the larger mass. It is 100 times harder to accelerate the smaller mass about its radius.

If the wheel is rotating at 0.5 radians/second, the 200 kg mass will be traveling at 0.01 m/s, and the 2 kg mass will be traveling 100 times faster at 1 m/s.

The 2 kg mass will have 100 times more kinetic energy than the larger mass, and 100 times more angular momentum. The kinetic energy came from the work required to accelerate the mass, therefore: the 2 kg mass required 100 times more work to accelerate it up to 0.5 radians per second.

And you seem to believe you can take linear momentum numbers from two different atwoods experiments, throw them into a third experiment and divide the linear momentum up evenly, even though as the 200 kg mass slows down to accelerate the 2 kg mass, it is 100 times harder to accelerate the 2 kg mass? That makes no sense whatsoever.

Yes, There are many ways to do it wrong. You demonstrated how applying linear momentum conservation to a rotating system can give you a bogus energy gain. In linear systems, the "mass-like-quantity" is just mass. In rotating systems, the "mass-like-quantity" is the moment of inertia (which you conveniently choose not to believe in).

[pequaide]

A one kilogram mass moving one meter per second on the end of a one meter string has a certain quantity of angular momentum. A one kilogram mass moving one meter per second on the end of a one hundred meter string has a different quantity of angular momentum. Cut both strings and the real momentum is the same. Therefore angular momentum is a false concept. I don't know if the blame goes on “moment of inertia� but the concept of angular momentum is false.

[Wubbly]

If you take a 1 kg mass and let it fall about 1.88 meters, it will take about 0.62 seconds to fall.

If you take the same mass attached to a very light rod 3 meters long, raise it 18 degrees from the horizontal and let it fall 36 degrees, it will swing through an arc length of 1.885 meters, and fall about 1.85 meters, in roughly 0.6 seconds.

If you double the rod length to 6 meters, and use a 9 degree initial angle with a 18 degree fall, it will swing through the same 1.885 meter arc length, and fall about 1.88 meters, in about 0.6 seconds.

If you triple the rod length to 9 meters, and use a 6 degree initial angle with a 12 degree fall, it will swing through the same 1.885 meter arc length, it will fall about 1.88 meters, in about 0.6 seconds.

In each case the mass travels the exact same distance (the same arc length), it falls roughly the same height (about 1.88 meters), and it takes roughly the same amount of time (about 0.6 seconds). You could set up photo gates, measure distance and time, and what would you conclude about these experiments?

Would you conclude that:

(A) They all fell about the same distance in about the same amount of time, therefore the length of the rod is irrelevant; You apply the same force to each mass therefore each mass is just as hard to move.

or would you conclude:

(B) In experiment 2 you applied twice the torque to go the same distance in roughly the same amount of time, therefore it was twice as hard to rotate the mass in experiment 2. In experiment 3 you applied three times the torque to go the same distance in roughly the same amount of time, therefore it was three times as hard to rotate the mass in experiment 3. Therefore: how hard it is to rotate the mass is directly proportional to the length of the rod.

or would you conclude something else?

[murilo]

Wubbly,
you are a lab guy!
Congrats!

If you use variable rods, or arms, with same weight, and START from the horizontal position, for sure you'll find equal 'g' acceleration operating at the edge.

From horizontal start the LAWS for pendulums are still not in operation!

The system sees the set as current free body fall under 'g'.

The pendulums laws - I mean behavior - will appear after the degree where a tangent will act just against the arm and the main holding fulcrum.

Then you'll deal to what is called angular velocity of oscillation, where differences will appear according elongations.

Best!
M

[Tarsier79]

It depends where the force is applied, vs where you measure it, and what reference frame you measure it in.

The stumbling block with force, is the force measurement doesn't account for velocity. Applying X force to a mass moving 1m/s takes less energy to apply the same force to a mass moving 2m/s.

[Wubbly]

In the above example the acceleration is due to gravity (9.81 m/s^2). The rod is considered massless, so you could say a force of roughly 9.81 N is being applied to the mass at the end of each rod. Each mass starts with zero velocity. Measure from a fixed cartesian coordinate system centered at the pivot point of each rod.

[eccentrically1]

I would conclude either (A) or something else.

The different lengths are irrelevant unless the pendulums are allowed to fall through more arc distance.
The force is the same so that has no bearing.

The photogates might reveal slight differences because of the longer lengths

[Wubbly]

I would be interested in pequaide's analysis of the falling mass/rod experiment.

Which of the following conclusions (if any), are correct?

(A) They all fell about the same distance in about the same amount of time, therefore the length of the rod is irrelevant; You apply the same force to each mass therefore each mass is just as hard to move.

or would you conclude:

(B) In experiment 2 you applied twice the torque to go the same distance in roughly the same amount of time, therefore it was twice as hard to rotate the mass in experiment 2. In experiment 3 you applied three times the torque to go the same distance in roughly the same amount of time, therefore it was three times as hard to rotate the mass in experiemnt 3. Therefore: As the mass gets farther away from the pivot point, it gets proportionally harder to rotate the mass (the increase is directly proportional to the length of the rod).


There is a third possible conclusion:

(C) In experiment 2 you applied twice the torque, but you only travelled half the angle, therefore it must have been FOUR times harder to rotate the mass about a point twice as far away. In experiment 3 you applied three times the torque, but you only travelled a third of the angle, therefore it must have been NINE times harder to rotate the mass about a point three times farther away. Therefore: As the mass gets farther away from the pivot point, it gets exponentially harder to rotate the mass.

(D) none of the above.
.

[pequaide]

Good thoughts Wubbly.

First: In your free fall with three rods and three arc thought experiment; the answer is. There is no torque. All three arc arrangements are freely falling. The three rods force the mass to fall around an arc but there is no resistance at the center bearing. The center bearing places no rotational force upon the mass and the mass places no rotational force upon the bearing. The arc makes it fall with less force in any position other than straight down, but the arc also gives the mass more distance over which the smaller force can act. It all works out that; if they fall the same distance the final velocity is always the same.

You have picked three arcs that are similar; so the times are similar. But the time differences are there. For proof of the time differences; set up the three rods in a pendulum arrangement with your arc going equal ways of 6 o'clock. The down swing velocity at 6 o'clock will be the same if the drop distances are equal; but the periods of the pendulums will be different.

Your three meter rod will cover the distance slower because the initial downward force is smaller. Note that the three meter rod does not drop the same distance. 1.85 is not are far as 1.88 and this is (unless this is a typo) what helps make the time shorter and closer to the other three arcs.

And your time is rounded way to much (.6), the photo gate can accurately measure 10 thousandths. The gates would pick up the time difference.

I pick conclusion (A)

Now lets go up to your 'wub-aa torque and angles.png' insert.

Your insert shows that torque is the same. Isn't torque the measure of the difficulty of rotating something: or the measure of force applied? Yet you say it is moment of inertia that is the measure of the difficulty of rotating something. You could put a pin at the center of rotation and make the lever arm into a top (toy). And it would spin quite nicely wouldn't it. The whole concept of balance means that one side rotates as easily as the other. Yet you say it is 100 times harder.

If you applied extra force to one side the 2 kg would rotate just as easily as the 200 kg side. So lets do it. Lets apply 1300 kilograms extra to one side at 2 cm from the pivot point. The 2 kilograms at 200 cm is like 200 kg at 2 cm so it is an Atwood's with 1300 kg accelerating 1700 kilograms 1300 kg /1700 kg * 9.81 m/sec/sec = 7.5 m/sec/sec. This will give you a velocity of 3.87 m/sec, at the 2 cm position, after a drop of 1 meter. This would mean that the 2 kilograms would be moving 387 m/sec. Humm; and we could replace the 200 kilograms at 2 cm with 2 kilograms at 200 cm and have 4 kilograms moving 387 m/sec. And we have only dropped 1.3 metric tons 1 meter; 12,753 joules. And 4 kilogram moving 387 m/sec is 299,538 joules.

Torque is torque balance means balanced momentum is conserved and F = ma is true It checks out in my physics book. And in all the experiments,

[pequaide]

This might help with the no torque rods.

Use a router to cut three deep grooves in metal and force your one kilogram mass to travel down the groove. Coat the mass in dry ice and forget about friction. Cut the three grooves with the same vertical height of 1.88 meters and with a radius of 3, 6, and 9 meters.

Now the mass runs down a groove but there can be no torque because there is no lever arm. The 3 meter radius groove is the longest distance traveled with the application of smaller forces; therefore it will take the most time. The 9 meter radius groove is the shortest distance (of the three) traveled and the forces are closer to straight down.

A straight down groove is like free fall and is shorter yet (the shortest) and all the forces applied are at max. 9.81 newtons the entire distance. All the other grooves apply some force to the side of the groove. The rod apply sideways force to the bearing but no rotational force is applied to the bearing. No torque

[Wubbly]

pequaide wrote:

...There is no torque. All three arc arrangements are freely falling. The three rods force the mass to fall around an arc but there is no resistance at the center bearing. ...

lol, You can't be serious. Torque is force x distance. There is a force acting on each mass. If you pick the start distance at the pivot point, then the torque is the force (m*g) times the rod length. (actually you need to multiply it by the sine or cosine of the angle depending on how you define your angles, but m*g is a good enough approximation for this experiment.

But the time differences are there

I understand the time differences are there. But I picked long rods in the horizontal position so the times would be roughly the same. The cosine of small angles is close to one (cos 15 = .966) so the force is about the same on all the masses. It's a rough approximation to demonstrate a point. If the rods were in the vertical position, the force on the masses would not be approximately the same.

I pick conclusion (A)

I thought you might pick that one. I picked conclusion (C).

Isn't torque the measure of the difficulty of rotating something.

You make me laugh. No. Torque is NOT the measure of difficulty of rotating something. Torque is a force times a distance (for example mass x acceleration-due-to-gravity x rod-length). Moment of inertia is the measure of difficulty of rotating something. They are two different rotational concepts. The mass/rod example was hopefully going to explain to you moment of inertia, but you can't even see the torque in the mass/rod experiment.

If you set up photo gates, measure distance and time, and calculate acceleration, and think linearly, you could prove F=ma, but you might not grasp the concept of moment of inertia because it deals with angular acceleration, not linear acceleration.